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Proof step 21 from proof bitsor-app (previous step)

# bits4 = # bits0, ∀ bits1, bits, bits2, bits3. # bits3 < #(bits4 + boolvar ') → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits = [], bits1 = [], bitsor(bits + bits4, bits1 + bits0) = bitsor(bits, bits1) + bitsor(bits4, bits0), bits4 = [], bits0 = []

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits, bits2, bits3] → [bits1, bits, bits0, bits4]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits1, bits, bits0, bits4